$C{H_3} - CH - C{H_2} - C{H_3}\xrightarrow[{UV}]{{C{l_2}}}A\,\left( {{\rm{major}}} \right),\,A$ is

$C{H_3} - \mathop {CH}\limits^{\mathop |\limits^{C{H_3}} } - \mathop {CH}\limits_{\mathop |\limits_{Cl} } - C{H_3}$

$C{H_3} - \mathop C\limits^{\mathop |\limits^{C{H_3}} } - C{H_2} - C{H_2}Cl$

$C{H_3} - \mathop {CH}\limits^{\mathop |\limits^{C{H_3}Cl} } - C{H_2} - C{H_3}$

NEET UG CHEMISTRY MCQ: HALOALKANES AND HALO ARENES

1 Medium

1 of 1

Application +4 | -1

C{H_3} - CH - C{H_2} - C{H_3}\xrightarrow[{UV}]{{C{l_2}}}A\,\left( {\mathrm{major}} \right),\,A

View answer and explanation

Correct option: B

C{H_3} - \mathop {\mathop C\limits_{\mathop |\limits_{Cl} } }\limits^{\mathop |\limits^{C{H_3}} } - C{H_2} - C{H_3}

Option B (

C{H_3} - \mathop {\mathop C\limits_{\mathop |\limits_{Cl} } }\limits^{\mathop |\limits^{C{H_3}} } - C{H_2} - C{H_3}

) is correct because chlorination of alkanes under UV light follows a free radical mechanism, and the most stable radical is formed at the most substituted carbon.

NEET UG CHEMISTRY MCQ: HALOALKANES AND HALO ARENES

More Questions from HALOALKANES AND HALO ARENES